Answer to Question #283898 in Mechanics | Relativity for straus

Question #283898

A 3500-kg car hits an 80-kg pedestrian who is standing in the middle of the road. Before the collision, the car was travelling at 30 km/h and the pedestrian was stationary. After the collision, the car was traveling at 28.5 km/h. At what speed will the pedestrian be flung down the road (in km/h)


1
Expert's answer
2022-01-02T18:29:04-0500

mass of Car (Mc)=3500kg(M_c)=3500kg

Mass of pedestrian(Mp)=80kg(M_p)=80kg

Initial velocity before collision


Vc=30km/hr=30×10003600m/sec=15018m/secvp=0km/hrV_c=30km/hr=\frac{30\times1000}{3600}m/sec=\frac{150}{18}m/sec\\v_p=0km/hr

After collision


Vc=28.5km/hr=28.5×10003600m/sec=1425180m/secV'_c=28.5km/hr=\frac{28.5\times1000}{3600}m/sec=\frac{1425}{180}m/sec

Momentum conservation

Momentum Before collision= momentum after collision

McVc+MpVp=McVc+MpVpM_cV_c+M_pV_p=M_cV'_c+M_pV'_p

Put value


3500×15018+80×0=3500×1425180+80×Vp3500\times\frac{150}{18}+80\times0=3500\times\frac{1425}{180}+80\times V'_p

29166.66=27708.33+80Vp29166.66=27708.33+80V'_p

1458.33=80VpVp=28.2291m/sec1458.33=80V'_p\\V'_p=28.2291m/sec


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